To find the domain of the function \( f(x) = \sqrt{4 - x^2} + \frac{1}{\sqrt{|\sin x| - \sin x}} \), we need to analyze the two components of the function separately.
### Step 1: Determine the domain of \( f_1(x) = \sqrt{4 - x^2} \)
For the square root function to be defined, the expression inside the square root must be non-negative:
\[
4 - x^2 \geq 0
\]
Rearranging this inequality gives:
\[
x^2 \leq 4
\]
Taking the square root of both sides, we find:
\[
-2 \leq x \leq 2
\]
Thus, the domain of \( f_1(x) \) is:
\[
D_1 = [-2, 2]
\]
### Step 2: Determine the domain of \( f_2(x) = \frac{1}{\sqrt{|\sin x| - \sin x}} \)
For this function to be defined, the expression inside the square root must be positive (since it is in the denominator):
\[
|\sin x| - \sin x > 0
\]
This inequality holds when \( \sin x < 0 \) because:
- If \( \sin x \) is negative, then \( |\sin x| = -\sin x \), and thus \( |\sin x| - \sin x = -\sin x - \sin x = -2\sin x > 0 \) implies \( \sin x < 0 \).
The sine function is negative in the intervals:
\[
(2n\pi - \pi, 2n\pi) \quad \text{for integers } n
\]
For our analysis, we will consider the first few intervals:
- For \( n = 0 \): \( (-\pi, 0) \)
- For \( n = 1 \): \( (2\pi - \pi, 2\pi) = (\pi, 2\pi) \)
- For \( n = -1 \): \( (-2\pi, -\pi) \)
### Step 3: Find the intersection of the domains
Now we need to find the intersection of the domains \( D_1 = [-2, 2] \) and the intervals where \( \sin x < 0 \).
From the intervals where \( \sin x < 0 \) within \( D_1 \):
- The interval \( (-\pi, 0) \) intersects with \( D_1 \) as \( [-2, 0) \).
- The interval \( (0, \pi) \) does not intersect with \( D_1 \) since \( 0 \) is not included.
Thus, the valid intersection is:
\[
D = [-2, 0)
\]
### Final Answer
The domain of \( f(x) \) is:
\[
[-2, 0)
\]
